Joseph Poveromo

25 Reputation

3 Badges

17 years, 60 days

MaplePrimes Activity


These are replies submitted by Joseph Poveromo

@Mariusz Iwaniuk  You are correct.  Thanks for pointing that out. The equation should read,

int[y'(x)*(x^2)/[(x^2)-1,x] = (int[sqrt(y(x))])^(-2/3)

Equation := int(diff(y(x), x)*w(x), x) = 1/int(sqrt(y(x))*w(x), x)^(2/3);
Solution := 3/4*y(x)^(4/3) + int(2/3*y(x)^(5/6)/int(sqrt(y(x))*w(x), x)^(5/3), x) + _C1 = 0;
Again, the advice provided by 'odeadvisor', was to formulate the Equation to the form
y(x) = G(x,y'(x)), then utilize the 'patterns' method which I could not apply, therefore it is the missing steps between 'Equation' and 'Solution'

Sorry,

JJP

Your advice is well taken - Thanks. Incorporating it, I will give it a second shot - I hope you will too. Equation := int(diff(y(x), x)*w(x), x) = 1/int(sqrt(y(x))*w(x), x)^(2/3); Solution := 3/4*y(x)^(4/3) + int(2/3*y(x)^(5/6)/int(sqrt(y(x))*w(x), x)^(5/3), x) + _C1 = 0; Again, the advice provided by 'odeadvisor', was to formulate the Equation to the form y(x) = G(x,y'(x)), then utilize the 'patterns' method which I could not apply, therefore it is the missing steps between 'Equation' and 'Solution' Thanks Again, JJP

Corection: Sorry all, I left out an exponential factor on the left side of the equation. 

The Equation should read,

 (2/3)*[int(diff(y(x),x)*x^2/(x^2 -1),x)]^(-3/2) =int(-sqrt(2*y(x)),x). It is that exponential

factor of (-3/2) on the left side of the equation which I left out...JJP

@Preben Alsholm 

Thanks for taking the time. It is who made the error. The Equation should have been (2/3)*[int(diff(y(x),x)*x^2/(x^2 -1),x)]^(-3/2) =int(-sqrt(2*y(x)),x). It is that exponential factor of (-3/2) on the left side of the equation which I left out. Sorry

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